easier. by observation the LHS is uniformly increasing with x. Since any positive value to the 0 power equals 1, by observation x must = 0, and no other real solution is possible.
In the real world, all the quantities of the left may only be positive. If we say f(x) = 2ˣ+ (⅔)ˣ+ (¾)ˣ then f(x) > 0 for all real x. f'(x) = 2ˣln 2 + (⅔)ˣln ⅔+ (¾)ˣln ¾ = 2ˣln 2 + (⅔)ˣ(ln 2- ln 3)+ (¾)ˣ(ln 3-2 ln 2) At x = 0, we find f'(0) = ln 2 + (ln 2- ln 3)+ (ln 3-2 ln 2) = ln 2 + ln 2- ln 3+ ln 3-2 ln 2 = 0 f(0) = 3 The second derivative at 0 is (ln 2)² + (ln ⅔)² + (ln ¾)² Since this is positive, the function only increases with positive or negative change in x. solution x = 0
Let f(x) be the left-hand side of this equation. It is easy to show that f(x) is concave up everywhere and has a minimal point at x=0 and f(0)=3.
*What about complex solutions?*
this
easier. by observation the LHS is uniformly increasing with x. Since any positive value to the 0 power equals 1, by observation x must = 0, and no other real solution is possible.
Nice
I have watched your video several times and still do not understand why ABC need be equal.
😊🎉👍👍👍🎉😊
In the real world, all the quantities of the left may only be positive. If we say
f(x) = 2ˣ+ (⅔)ˣ+ (¾)ˣ
then f(x) > 0 for all real x.
f'(x) = 2ˣln 2 + (⅔)ˣln ⅔+ (¾)ˣln ¾
= 2ˣln 2 + (⅔)ˣ(ln 2- ln 3)+ (¾)ˣ(ln 3-2 ln 2)
At x = 0, we find
f'(0) = ln 2 + (ln 2- ln 3)+ (ln 3-2 ln 2)
= ln 2 + ln 2- ln 3+ ln 3-2 ln 2
= 0
f(0) = 3
The second derivative at 0 is
(ln 2)² + (ln ⅔)² + (ln ¾)²
Since this is positive, the function only increases with positive or negative change in x.
solution
x = 0
Nice!